The velocity of ball before entering the water surface

$\mathbf{v} = \sqrt{2gh} = \sqrt{2g \times 9}$

When ball enters into water, due to up thrust of water the velocity of ball decreases (or retarded)

$The retardation, a = \frac{\text{apparent weight}}{\text{mass of ball}} = \frac{V(\rho - \sigma)g}{V\rho} = \left(\frac{\rho - \sigma}{\rho}\right) g = \left(\frac{0.4 - 1}{0.4}\right) \times g = -\frac{3}{2} g If h be the depth up to which ball sink, then, 0 - v^2 = 2 \times \left(-\frac{3}{2} g\right) \times h \Rightarrow 2g \times 9 = 3gh \therefore h = 6 \text{ cm}.$